Algebra Examples

$\frac{x}{x^{2} + 1} \leq 1$

Step 1

Subtract $1$ from both sides of the inequality.

$\frac{x}{x^{2} + 1} - 1 \leq 0$

Step 2

Simplify $\frac{x}{x^{2} + 1} - 1$ .

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Step 2.1

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{x^{2} + 1}{x^{2} + 1}$ .

$\frac{x}{x^{2} + 1} - 1 \cdot \frac{x^{2} + 1}{x^{2} + 1} \leq 0$

Step 2.2

Combine $- 1$ and $\frac{x^{2} + 1}{x^{2} + 1}$ .

$\frac{x}{x^{2} + 1} + \frac{- (x^{2} + 1)}{x^{2} + 1} \leq 0$

Step 2.3

Combine the numerators over the common denominator.

$\frac{x - (x^{2} + 1)}{x^{2} + 1} \leq 0$

Step 2.4

Simplify the numerator.

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Step 2.4.1

Apply the distributive property.

$\frac{x - x^{2} - 1 \cdot 1}{x^{2} + 1} \leq 0$

Step 2.4.2

Multiply $- 1$ by $1$ .

$\frac{x - x^{2} - 1}{x^{2} + 1} \leq 0$

Step 2.4.3

Reorder terms.

$\frac{- x^{2} + x - 1}{x^{2} + 1} \leq 0$

Step 2.5

Factor $- 1$ out of $- x^{2}$ .

$\frac{- (x^{2}) + x - 1}{x^{2} + 1} \leq 0$

Step 2.6

Factor $- 1$ out of $x$ .

$\frac{- (x^{2}) - 1 (- x) - 1}{x^{2} + 1} \leq 0$

Step 2.7

Factor $- 1$ out of $- (x^{2}) - 1 (- x)$ .

$\frac{- (x^{2} - x) - 1}{x^{2} + 1} \leq 0$

Step 2.8

Rewrite $- 1$ as $- 1 (1)$ .

$\frac{- (x^{2} - x) - 1 (1)}{x^{2} + 1} \leq 0$

Step 2.9

Factor $- 1$ out of $- (x^{2} - x) - 1 (1)$ .

$\frac{- (x^{2} - x + 1)}{x^{2} + 1} \leq 0$

Step 2.10

Rewrite $- (x^{2} - x + 1)$ as $- 1 (x^{2} - x + 1)$ .

$\frac{- 1 (x^{2} - x + 1)}{x^{2} + 1} \leq 0$

Step 2.11

Move the negative in front of the fraction.

$- \frac{x^{2} - x + 1}{x^{2} + 1} \leq 0$

Step 3

Find all the values where the expression switches from negative to positive by setting each factor equal to $0$ and solving.

$x^{2} - x + 1 = 0$

$x^{2} + 1 = 0$

Step 4

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 5

Substitute the values $a = 1$ , $b = - 1$ , and $c = 1$ into the quadratic formula and solve for $x$ .

$\frac{1 \pm \sqrt{{(- 1)}^{2} - 4 \cdot (1 \cdot 1)}}{2 \cdot 1}$

Step 6

Simplify.

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Step 6.1

Simplify the numerator.

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Step 6.1.1

Raise $- 1$ to the power of $2$ .

$x = \frac{1 \pm \sqrt{1 - 4 \cdot 1 \cdot 1}}{2 \cdot 1}$

Step 6.1.2

Multiply $- 4 \cdot 1 \cdot 1$ .

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Step 6.1.2.1

Multiply $- 4$ by $1$ .

$x = \frac{1 \pm \sqrt{1 - 4 \cdot 1}}{2 \cdot 1}$

Step 6.1.2.2

Multiply $- 4$ by $1$ .

$x = \frac{1 \pm \sqrt{1 - 4}}{2 \cdot 1}$

Step 6.1.3

Subtract $4$ from $1$ .

$x = \frac{1 \pm \sqrt{- 3}}{2 \cdot 1}$

Step 6.1.4

Rewrite $- 3$ as $- 1 (3)$ .

$x = \frac{1 \pm \sqrt{- 1 \cdot 3}}{2 \cdot 1}$

Step 6.1.5

Rewrite $\sqrt{- 1 (3)}$ as $\sqrt{- 1} \cdot \sqrt{3}$ .

$x = \frac{1 \pm \sqrt{- 1} \cdot \sqrt{3}}{2 \cdot 1}$

Step 6.1.6

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \frac{1 \pm i \sqrt{3}}{2 \cdot 1}$

Step 6.2

Multiply $2$ by $1$ .

$x = \frac{1 \pm i \sqrt{3}}{2}$

Step 7

The final answer is the combination of both solutions.

$x = \frac{1 + i \sqrt{3}}{2}, \frac{1 - i \sqrt{3}}{2}$

Step 8

Subtract $1$ from both sides of the equation.

$x^{2} = - 1$

Step 9

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{- 1}$

Step 10

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \pm i$

Step 11

The complete solution is the result of both the positive and negative portions of the solution.

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Step 11.1

First, use the positive value of the $\pm$ to find the first solution.

$x = i$

Step 11.2

Next, use the negative value of the $\pm$ to find the second solution.

$x = - i$

Step 11.3

The complete solution is the result of both the positive and negative portions of the solution.

$x = i, - i$

Step 12

Solve for each factor to find the values where the absolute value expression goes from negative to positive.

$x = \frac{1 + i \sqrt{3}}{2}, \frac{1 - i \sqrt{3}}{2}$

$x = i, - i$

Step 13

The leading coefficient cannot be determined because $- \frac{x^{2} - x + 1}{x^{2} + 1}$ is not a polynomial.

Not a polynomial

Step 14

Since there are no real x-intercepts and the leading coefficient is positive, the parabola opens up and $- \frac{x^{2} - x + 1}{x^{2} + 1}$ is always greater than $0$ .

No solution